Q. A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm, the ice-cream melts at the rate of $81\ \mathrm{cm^3/min}$ and the thickness of the ice-cream layer decreases at the rate of $\frac1{4\pi}$ cm/min. The surface area (in $\mathrm{cm^2}$) of the chocolate ball (without the ice-cream layer) is :
A) $196\pi$
B) $256\pi$
C) $225\pi$
D) $128\pi$
Explanation
[Solution by Dr. Abubacker] Let $r$ be the ball's radius and $R=r+1$ the outer radius. $\frac{dV}{dt}=4\pi R^2\frac{dR}{dt}$: $81=4\pi R^2\cdot\frac1{4\pi}\Rightarrow R=9$, so $r=8$. Surface area $=4\pi r^2=256\pi$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.